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Let $G$ be a first countable Hausdorff topological group and $g\in G\setminus \{e\}$. Denote $[g]$ be the conjugacy class of $g$.

My question is that "Can $e\in \overline{[g]}$?", where $\overline{[g]}$ means the closure of $[g]$ in $G$.

Equivalently, "Does there exist a sequence $\{x_n\}$ in $G$ such that $x_n^{-1}gx_n\to e$ as $n\to \infty$?"

Or, we can just consider more explicit case that $G$ assume to be second countable locally compact Hausdorff group.

We already know that the answer is NO when the group $G$ is abelian or compact or discrete. But we can't find a solution for general cases.

Thanks in advance.

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    $\begingroup$ I believe there are certainly cases where this is true. The conjugacy class of the matrix $$\begin{pmatrix} 1&1 \\ 0&1\end{pmatrix}$$ contains the identity in its closure (if over at least $\mathbb{Q}$). $\endgroup$ – Santana Afton May 21 '18 at 13:17

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